Proposition 2

Let $(\mathfrak{g}_i)_{i\in I}$ be a finite family of Lie algebras and $\mathfrak{g}=\prod_{i\in I}\mathfrak{g}_i$.

The Cartan subalgebras of $\mathfrak{g}$ are the subalgebras of the form $\prod_{i\in I}\mathfrak{h}_i$, where $\mathfrak{h}_i$

is a Cartan subalgebra of $\mathfrak{g}_i$.

If $\mathfrak{h}_i$ is a subalgebra of $\mathfrak{g}_i$ with normalizer $\mathfrak{n}_i$, then $\prod\mathfrak{h}_i$ is a subalgebra of $\mathfrak{g}$ with normalizer $\prod\mathfrak{n}_i$; if the $\mathfrak{h}_i$ are nilpotent, $\prod\mathfrak{h}_i$ is nilpotent; thus, if $\mathfrak{h}_i$ is a Cartan subalgebra of $\mathfrak{g}_i$ for all $i,\prod\mathfrak{h}_i$ is a Cartan subalgebra of $\mathfrak{g}$. Conversely, let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$; the projection $\mathfrak{h}_i$ of $\mathfrak{h}$ onto $\mathfrak{g}_i$ is a nilpotent subalgebra of $\mathfrak{g}_i$, and $\prod\mathfrak{h}_i$ is a nilpotent subalgebra of $\mathfrak{g}$ containing $\mathfrak{h}$; hence $\mathfrak{h}=\prod\mathfrak{h}_i$ (Prop. 1); thus, for all $i,\mathfrak{h}_i$ is its own normalizer in $\mathfrak{g}_i$, and so is a Cartan subalgebra of $\mathfrak{g}_i$.